Explanation:
given that AD > DC, we let AD = kDC, where k>1. for simplicity's sake we let DC = x and BD = y. so AD is kx.
we can find AB using the Pythagorean theorem.
AB^2=k^2x^2+y^2.
same with BC.
BC^2=x^2+y^2
now compare AB^2 and BC^2. obviously here AB^2 > BC^2. let AB^2 = mBC^2, where m>1
we then take the square roots of these
AB = sqrt(m)BC, and sqrt(m)>1.
thus, we have AB = nBC, where n=sqrt(m).
therefore, AB>BC.